Bài 1:

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=>a=bk; c=dk

\(\frac{5a+4b}{5a-4b}=\frac{5\cdot bk+4b}{5\cdot bk-4b}=\frac{b\left(5k+4\right)}{b\left(5k-4\right)}=\frac{5k+4}{5k-4}\)

\(\frac{5c+4d}{5c-4d}=\frac{5\cdot dk+4d}{5\cdot dk-4d}=\frac{d\left(5k+4\right)}{d\left(5k-4\right)}=\frac{5k+4}{5k-4}\)

Do đó: \(\frac{5a+4b}{5a-4b}=\frac{5c+4d}{5c-4d}\)

Bài 2:

a: Đặt \(\frac{a}{b}=\frac{b}{c}=k\)

=>b=ck; a=bk=ck^2

\(\frac{a^2+b^2}{b^2+c^2}=\frac{\left(ck^2\right)^2+\left(ck\right)^2}{\left(ck\right)^2+c^2}=\frac{c^2k^2\left(k^2+1\right)}{c^2\left(k^2+1\right)}=k^2\)

\(\frac{a}{c}=\frac{ck^2}{c}=k^2\)

Do đó: \(\frac{a}{c}=\frac{a^2+b^2}{b^2+c^2}\)

b: Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\)

=>\(\begin{cases}c=dk\\ b=ck=dk\cdot k=dk^2\\ a=bk=dk^2\cdot k=dk^3\end{cases}\)

\(\left(\frac{a+b-c}{b+c-d}\right)^3=\left(\frac{dk^3+dk^2-dk}{dk^2+dk-d}\right)^3\)

\(=\left\lbrack\frac{dk\left(k^2+k-1\right)}{d\left(k^2+k-1\right)}\right\rbrack^3=k^3\)

\(\frac{a}{d}=\frac{dk^3}{d}=k^3\)

Do đó: \(\left(\frac{a+b-c}{b+c-d}\right)^3=\frac{a}{d}\)

Ta có:2bd=c(b+d)

=>2bd=bc+cd

Mà a+c=2b (theo đề)

=>(a+c).d=bc+cd

=>ad+cd=bc+cd

=>ad=bc (cùng bớt đi cd)

=>a/b=c/d (đpcm)

a. Áp dụng tính chất dãy tỉ số = nhau ta có:
x/4 = y/3 = z/9 = (x - 3y + 4z)/(4 - 9 + 36) = 62/31 = 2
=> x/4 = 2 -> x = 8
     y/3 = 2 -> y = 6
     z/9 = 2 -> z = 18

cứ làm đi 3 con tích sẽ về ngay tay bn

Bài 1:

G/s ngược lại: \(ad=bc\) , ta cần CM giả thiết.

Ta có: \(ad=bc\) => \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\) \(\left(k\inℤ\right)\)

Thay vào:

\(\left(a+b+c+d\right)\left(a-b-c+d\right)\)

\(=\left(bk+b+dk+d\right)\left(bk-b-dk+d\right)\)

\(=\left(k+1\right)\left(b+d\right)\left(k-1\right)\left(b-d\right)\) (1)

\(\left(a-b+c-d\right)\left(a+b-c-d\right)\)

\(=\left(bk-b+dk-d\right)\left(bk+b-dk-d\right)\)

\(=\left(k-1\right)\left(b+d\right)\left(k+1\right)\left(b-d\right)\) (2)

Từ (1) và (2) => GT được CM => đpcm

Ta có :

\(\frac{a}{b}< \frac{c}{d}\Leftrightarrow ad< ac\Leftrightarrow ab+ad< ab+bc\Leftrightarrow a\left(b+d\right)< b\left(a+c\right)\)\(\Leftrightarrow\frac{a}{b}< \frac{a+c}{b+d}\left(1\right)\)

\(\frac{a}{b}< \frac{c}{d}\Leftrightarrow bc>ad\Leftrightarrow bc+cd>ad+cd\)\(\Leftrightarrow c\left(b+d\right)>d\left(a+c\right)\Leftrightarrow\frac{c}{d}>\frac{a+c}{b+d}\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\)

\(\frac{a}{b}< \frac{c}{d}\left(1\right).\)Nhân 2 vế của (1) với bd ta có:

\(\frac{a}{b}\times bd=ad< \frac{c}{d}\times bd=bc\)( đpcm )

ad < bc ( 2 ).Chia 2 vế của (2) cho bd ta có:

\(\frac{ad}{bd}=\frac{a}{b}< \frac{bc}{bd}=\frac{c}{d}\left(Đpcm\right)\)

(a+b+c+d)(a-b-c+d)=(a-b+c-d)(a+b-c-d)

=>\(\left(a+d\right)^2-\left(b+c\right)^2=\left(a-d\right)^2-\left(b-c\right)^2\)

=>\(\left(a+d\right)^2-\left(a-d\right)^2=\left(b+c\right)^2-\left(b-c\right)^2\)

=>(a+d-a+d)(a+d+a-d)=(b+c-b+c)(b+c+b-c)

=>\(2d\cdot2a=2c\cdot2b\)

=>ad=bc

=>\(\frac{a}{c}=\frac{b}{d}\)